In fig, $A B C$ is a right triangle right angled at $A . B C E D, A C F G$ and $A B M N$ are square on the sides $B C, C A$ and $A B$ respectively. Line segment $A X ⊥ D E$ meets $B C$ at $Y$ . Show that:
(i) $△ M B C ≅ △ A B D$
(ii) $a r (B Y X D) = 2 a r (M B C)$
(iii) $a r (B Y X D) = a r (A B M N)$
(iv) $△ F C B ≅ △ A C E$
(v) $a r (C Y X E) = 2 a r (F C B)$
(vi) $a r (C Y X E) = a r (A C F G)$
(vii) $a r (B C E D) = a r (A B M N) + a r (A C F G)$
Result (vii) is the famous Theorem of Pythagoras. You shall learn a simpler
proof of this theorem in Class X.

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Solution

(i)
In $Δ s M B C$ and $A B D$ , we have
$B C = B D$
[Sides of the square BCED]
$M B = A B$
[Sides of the square ABMN]
$∠ M B C = ∠ A B D$
[Since Each $= 90 º + ∠ A B C$ ]
Therefore by SAS criterion of congruence, we have
$Δ M B C ≅ δ A B D$
(ii)
$Δ A B D$ and square $B Y X D$ have the same base $B D$ and are between the same parallels $B D$ and $A X$ .
Therefore $a r (A B D) = \frac{1}{2} a r (B Y X D)$
But $Δ M B C ≅ Δ A B D$ [Proved in part (i)]

$\Rightarrow a r (M B C) = a r (A B D)$
Therefore $a r (M B C) = a r (A B D) = \frac{1}{2} a r (B Y X D)$
$\Rightarrow a r (B Y X D) = 2 a r (M B C) .$
(iii)
Square $A B M N$ and $Δ M B C$ have the same base $M B$ and are between same parallels $M B$ and $N A C$ .
Therefore $a r (M B C) = \frac{1}{2} a r (A B M N)$
$\Rightarrow a r (A B M N) = 2 a r (M B C)$
$= a r (B Y X D)$ [Using part (ii)]
(iv)
In $Δ s$ ACE and BCF, we have
$C E = B C$ [Sides of the square BCED]
$A C = C F$ [Sides of the square ACFG]
and $∠ A C E = ∠ B C F$ [Since Each $= 90 º + ∠ B C A$ ]
Therefore by SAS criterion of congruence,
$Δ A C E ≅ Δ B C F$
(v)
$Δ A C E$ and square $C Y X E$ have the same base $C E$ and are between same parallels $C E$ and $A Y X$ .
Therefore $a r (A C E) = \frac{1}{2} a r (C Y X E)$
$\Rightarrow a r (F C B) = \frac{1}{2} a r (C Y X E)$ [Since $Δ A C E ≅ Δ B C F$ , part (iv)]
$\Rightarrow a r (C Y X E) = 2 a r (F C B)$ .
(vi)
Square $A C F G$ and $\Rightarrow B C F$ have the same base $C F$ and are between same parallels $C F$ and $B A G$ .
Therefore $a r (B C F) = \frac{1}{2} a r (A C F G)$
$\Rightarrow \frac{1}{2} a r (C Y X E) = \frac{1}{2} a r (A C F G)$ [Using part (v)]
$\Rightarrow a r (C Y X E) = a r (A C F G)$
(vii)
From part (iii) and (vi) we have
$a r (B Y X D) = a r (A B M N)$
and
$a r (C Y X E) = a r (A C F G)$
On adding we get
$a r (B Y X D) + a r (C Y X E) = a r (A B M N) + a r (A C F G) a r (B C E D) = a r (A B M N) + a r (A C F G)$

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Similar questions

In the following figure, ABC is a right triangle right angled at A. BCED, ACFG and ABMN are squares on the sides BC, CA and AB respectively. Line segment AX ⊥ DE meets BC at Y. Show that:

(i) ΔMBC ≅ ΔABD

(ii)

(iii)

(iv) ΔFCB ≅ ΔACE

(v)

(vi)

(vii)

Note: Result (vii) is the famous Theorem of Pythagoras. You shall learn a simpler proof of this theorem in class X.

Question 8
In the following figure, ABC is a right triangle right angled at A. BCED, ACFG and ABMN are squares on the sides BC, CA and AB respectively. Line segment $A X ⊥ D E$ meets BC at Y. Show that:

(i) $Δ M B C ≅ Δ A B D$
(ii) $a r e a (B Y X D) = 2 a r e a (Δ M B C)$
(iii) $a r (B Y X D) = a r (A B M N)$
(iv) $Δ F C B ≅ Δ A C E$
(v) $a r (C Y X E) = 2 a r (Δ F C B)$
(vi) $a r (A C F G) = a r (C Y X E)$
(vii) ar (BCED) = ar(ABMN) + ar(ACFG)

Question 8
In the following figure, ABC is a right triangle right angled at A. BCED, ACFG and ABMN are squares on the sides BC, CA and AB respectively. Line segment $A X ⊥ D E$ meets BC at Y. Show that:

(i) $Δ M B C ≅ Δ A B D$
(ii) $a r e a (B Y X D) = 2 a r e a (Δ M B C)$
(iii) $a r (B Y X D) = a r (A B M N)$
(iv) $Δ F C B ≅ Δ A C E$
(v) $a r (C Y X E) = 2 a r (Δ F C B)$
(vi) $a r (A C F G) = a r (C Y X E)$
(vii) ar (BCED) = ar(ABMN) + ar(ACFG)

In the figure, ABC is a right triangle right angled at A. BCED, ACFG and ABMN are squares on the sides BC, CA and AB respectively. Line segment $A X ⊥ D E$ meets BC at Y Show that
$(i) △ M B C ≅ △ A B D (i i) a r (B Y X D) = 2 a r (△ M B C) (i i i) a r (B Y X D) = a r (A B M N) (i v) △ F C B ≅ △ A C E (v) a r (C Y X E) = 2 a r (△ F C B) (v i) a r (C Y X E) = a r (A C F G) (v i i) a r (B C E D) = a r (A M B N) + a r (A C F G)$

Q. If the given figure, ABC is a right triangle right angled at A, BCED, ACFGand ABMN are squares on the sides BC, CA and AB respectively. Line segment AX ⊥ DE meets BC at Y.

Show that
(i) Δ MBC

≅

Δ ABD

(ii) ar (BYXD) = 2 ar (Δ MBC)

(iii) ar (BYXD) = ar (ABMN)

(iv) Δ FCB

≅

Δ ACE

(v) ar (CYXE) = 2 ar (ΔFCB)

(vi) ar (CYXE) = ar (ACFG)

(vii) ar (BCED) = ar (ABMN) + ar (ACFG)

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